Exercise 10.4 · Q1: $|\vec a\times\vec b|$ for $\hat i-7\hat j+7\hat k$⁠, $3\hat i-2\hat j+2\hat k$

Basics you need

Solution

1. Set up the determinant

   $$\vec a\times\vec b = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -7 & 7 \\ 3 & -2 & 2 \end{vmatrix}.$$
   The second row holds the components of $\vec a$ (the first vector) and the third row those of $\vec b$⁠. The $\hat i$ part of $\vec a$ has coefficient 1, so the row starts with 1; the minus signs of $-7\hat j$ and $-2\hat j$ go into the entries.
   The determinant form of $\vec a\times\vec b$ (Section 10.6.3). Lesson: the cross product

2. Expand along the top row

   $$\begin{aligned} \vec a\times\vec b &= \hat i\begin{vmatrix} -7 & 7 \\ -2 & 2 \end{vmatrix} - \hat j\begin{vmatrix} 1 & 7 \\ 3 & 2 \end{vmatrix} \\ &\quad + \hat k\begin{vmatrix} 1 & -7 \\ 3 & -2 \end{vmatrix} \\ &= \hat i(-14 + 14) - \hat j(2 - 21) \\ &\quad + \hat k(-2 + 21) \\ &= 0\hat i + 19\hat j + 19\hat k. \end{aligned}$$
   Cover the row and column of each top entry and take the 2 by 2 determinant that is left: for $\hat i$⁠, $(-7)(2) - (7)(-2) = -14 + 14 = 0$⁠; for $\hat j$⁠, $(1)(2) - (7)(3) = 2 - 21 = -19$⁠; for $\hat k$⁠, $(1)(-2) - (-7)(3) = -2 + 21 = 19$⁠. The middle term carries a minus sign, so $-(-19) = +19$⁠. The $\hat i$ part is 0 because the last two components of $\vec a$⁠, $-7$ and 7, are in the same ratio as those of $\vec b$⁠, $-2$ and 2.
   Expansion along the first row, signs $+, -, +$⁠. Basics: a determinant measures a signed area

3. Check: the answer is perpendicular to both vectors

   $$\begin{aligned} (\vec a\times\vec b)\cdot\vec a &= 0 - 133 + 133 = 0, \\ (\vec a\times\vec b)\cdot\vec b &= 0 - 38 + 38 = 0. \end{aligned}$$
   The products are $(0)(1) = 0$⁠, $(19)(-7) = -133$⁠, $(19)(7) = 133$ for $\vec a$⁠, and $(0)(3) = 0$⁠, $(19)(-2) = -38$⁠, $(19)(2) = 38$ for $\vec b$⁠. By definition $\vec a\times\vec b$ is perpendicular to both $\vec a$ and $\vec b$⁠, so both scalar products must be 0; a slip in a minor almost always breaks one of them.
   $\vec u\cdot\vec v = 0$ for perpendicular vectors (Section 10.6.1, Observation 2). Lesson: the dot product

4. The length

   $$\begin{aligned} |\vec a\times\vec b| &= \sqrt{0^2 + 19^2 + 19^2} \\ &= \sqrt{361 + 361} = \sqrt{722}. \end{aligned}$$

   Since $722 = 361 \times 2 = 19^2 \times 2$⁠, $\sqrt{722} = 19\sqrt2$⁠.

   Squaring each component and adding gives the square of the length. Taking out the square factor $19^2$ leaves the simplest surd form, the form the answer key prints.
   $|\vec r| = \sqrt{x^2 + y^2 + z^2}$ (Section 10.5.1). Basics: use Pythagoras twice

5. What $19\sqrt2$ measures, and a second check

   $|\vec a\times\vec b| = |\vec a||\vec b|\sin\theta$ is the area of the parallelogram with sides $\vec a$ and $\vec b$⁠. Squaring and using $\sin^2\theta = 1 - \cos^2\theta$⁠:

   $$\begin{aligned} |\vec a\times\vec b|^2 &= |\vec a|^2|\vec b|^2 - (\vec a\cdot\vec b)^2 \\ &= (99)(17) - 31^2 \\ &= 1683 - 961 = 722, \end{aligned}$$

   because $|\vec a|^2 = 1 + 49 + 49 = 99$⁠, $|\vec b|^2 = 9 + 4 + 4 = 17$ and $\vec a\cdot\vec b = 3 + 14 + 14 = 31$⁠.

   This second route uses only the scalar product, so it checks the determinant work independently. The identity comes from $|\vec a\times\vec b|^2 = |\vec a|^2|\vec b|^2\sin^2\theta$ and $(\vec a\cdot\vec b)^2 = |\vec a|^2|\vec b|^2\cos^2\theta$⁠.
   Area of a parallelogram $= |\vec a\times\vec b|$ (Section 10.6.3, Observation 9). Lesson: Examples 22–25

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